Question

the percentage purity of calcium carbonate sample whose 10 gram on complete decomposition gives 4 gram CO2 is​

Answer 1

Explanation:

Correct option is B)

CaCO3⇌ΔCaO+CO2

Let pure sample of CaCO3=x grams

Mass of CaO produced after decomposition =22.4 g

Molar mass of CaCO3=100 g/mol

and, Molar mass of CaO=56 g

If 100% is pure, then

100 g CaCO3⟶56 g of CaO

Also,y g of CaCO3⟶22.4 g of CaO

Dividing these two,

y100=22.456

y=40 grams

∴ Percentage of purity =TotalmassofimpureMassofpuresample×100=5040×100=80%

Answer 2

Answer:

The percentage purity of 10g of calcium carbonate sample  on complete decomposition gives 4 gram CO₂ is​ 90.9%.

Explanation:

The chemical equation of decomposition of calcium carbonate:-

CaCO₃ $\longrightarrow$  CaO  +  CO₂

In the given sample, the amount of calcium carbonate = 10g

The amount of carbon dioxide produced = 4g

We know that the molar mass of the calcium carbonate = 100g/mol

The molar mass of the carbon dioxide = 44g/mol

The number of moles of calcium carbonate  $=\frac{10}{100} =0.1 mol$

The number of moles of CO₂ $=\frac{4}{44}=0.0909\; mol$

From the chemical equation, one mole of CaCO₃ produces one mole of CO₂.

Therefore, 0.0909 moles of CO₂ will be produce by CaCO₃ = 0.0909 mol

The weight of 0.0909 mol of CaCO₃  $= 0.0909\times 100 =9.09g$

Percentage purity = (mass of pure chemical/ mass of sample)×100

Percentage purity of calcium carbonate sample $=\frac{9.09}{10} \times 100 =90.0\%$

Therefore, percentage purity of calcium carbonate sample is 90.0%.

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